Question

In fig. 9 is shown a right circular cone of height 30 cm. A small cone is cut off from the top by a plane parallel to the base. If the volume of the small cone is 127127 of the volume of cone, find at what height above the base is the section made.

Answer 1

Let the radius and height of the bigger cone be R and H, respectively.
Given: H = 30 cm
Let the radius and height of the smaller cone be r and h, respectively.

Now, in ∆AFE and ∆AGC,

∠AEF = ∠ACG       (Corresponding angles)
∠AFE = ∠AGC       (90°° each)

∴∆AFE~∆AGC      (AA similarity)

\[\Rightarrow \frac{AF}{AG} = \frac{FE}{GC}\]
\[ \Rightarrow \frac{h}{H} = \frac{r}{R} . . . . . \left( 1 \right)\]

It is given that
Volume of the smaller cone \[\frac{1}{27} \times\] Volume of the bigger cone    

\[\Rightarrow \frac{1}{3}\pi r^2 h = \frac{1}{27} \times \frac{1}{3}\pi R^2 H\]
\[ \Rightarrow \left( \frac{r}{R} \right)^2 \times \frac{h}{H} = \frac{1}{27}\]
\[ \Rightarrow \left( \frac{h}{H} \right)^2 \times \frac{h}{H} = \frac{1}{27} \left[ Using \left( 1 \right) \right]\]
\[ \Rightarrow \left( \frac{h}{H} \right)^3 = \frac{1}{27}\]
\[ \Rightarrow \frac{h}{H} = \frac{1}{3}\]

\[\therefore h = \frac{1}{3} \times H = \frac{1}{3} \times 30 = 10 cm\]

Now,
FG = AG − AF = 30 cm − 10 cm = 20 cm
Hence, the section is made 20 cm above the base.